# Find the values of a, b, c and k.

#### anemone

##### MHB POTW Director
Staff member
Hi,

The problem stated below has me stumped. I've been trying to solve it for some times but I kept ended up with zero equals zero kind of silly result and I'd appreciate it if someone could show me some ideas on how to crack it.

Problem:

Let $$\displaystyle a,\;b,\;c$$ be roots of equation $$\displaystyle x^3-6x^2+kx+k=0$$ and $$\displaystyle (a-1)^3+(b-2)^3+(c-3)^3=0$$. Find a, b, c and k.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Hi,

The problem stated below has me stumped. I've been trying to solve it for some times but I kept ended up with zero equals zero kind of silly result and I'd appreciate it if someone could show me some ideas on how to crack it.

Problem:

Let $$\displaystyle a,\;b,\;c$$ be roots of equation $$\displaystyle x^3-6x^2+kx+k=0$$ and $$\displaystyle (a-1)^3+(b-2)^3+(c-3)^3=0$$. Find a, b, c and k.
Since $a,b,c$ are roots of $x^3-6x^2+kx+k=0$, we have $a+b+c=6\Rightarrow (a-1)+(b-2)+(c-3)=0$. This means $(a-1)^3+(b-2)^3+(c-3)^3=3(a-1)(b-2)(c-3)$. So you get $(a-1)(b-2)(c-3)=0$.
Now
We already have $ab+bc+ca=k$ and $abc=-k$. This should help.

#### anemone

##### MHB POTW Director
Staff member
Since $a,b,c$ are roots of $x^3-6x^2+kx+k=0$, we have $a+b+c=6\Rightarrow (a-1)+(b-2)+(c-3)=0$. This means $(a-1)^3+(b-2)^3+(c-3)^3=3(a-1)(b-2)(c-3)$.
Thanks caffeinemachine for the quick reply..

I'm astonished to see another relation exists between $a,b,c$, i.e $(a-1)+(b-2)+(c-3)=0$. I must thank you for that! If I may start from $(a-1)+(b-2)+(c-3)=0$, I get:

$(a-1)+(b-2)=-(c-3)$

$((a-1)+(b-2))^3=(-(c-3))^3$

$(a-1)^3+3(a-1)^2(b-2)+3(a-1)(b-2)^2+(b-2)^3=-(c-3)^3$

$(a-1)^3++(b-2)^3+(c-3)^3+3(a-1)(b-2)((a-1)+(b-2))=0$

$(a-1)^3++(b-2)^3+(c-3)^3+3(a-1)(b-2)(a+b-3)=0$

$(a-1)^3++(b-2)^3+(c-3)^3+3(a-1)(b-2)(6-c-3)=0$

$(a-1)^3++(b-2)^3+(c-3)^3+3(a-1)(b-2)(3-c)=0$

Since we're told that $(a-1)^3++(b-2)^3+(c-3)^3=0$, it means $3(a-1)(b-2)(3-c)=0$, or

$a=1,\;b=2,\;c=3$.

Since $ab+bc+ac=k$ and $abc=-k$, we noticed that $a=1,\;b=2,\;c=3$ can't be true at the same time. In other words, we need to investigate for each of the cases.

Case 1 (When $$\displaystyle a=1$$):

From the Newton Identities we obtained the following equations:

$$\displaystyle a+b+c=6$$

$$\displaystyle a^2+b^2+c^2=36-2k$$

$$\displaystyle a^3+b^3+c^3=216-21k$$

Substitute $$\displaystyle a=1$$ into each of the equations above leads us to

$$\displaystyle b+c=5$$

$$\displaystyle b^2+c^2=35-2k$$

$$\displaystyle b^3+c^3=215-21k$$

And by using the identity $$\displaystyle b^3+c^3=(b+c)(b^2-bc+b^2)$$ and $$\displaystyle (1)(bc)=-k$$, we get:

$$\displaystyle k=\frac{5}{2},\;b=\frac{5-\sqrt{35}}{2},\;c=\frac{5+\sqrt{35}}{2}$$

And we repeat all these for the second and third cases to find

$$\displaystyle a=\frac{2(3+\sqrt{5})}{2},\;b=2,\;c=\frac{2(3-\sqrt{5})}{2},\;k=\frac{16}{3}$$

and

$$\displaystyle a=\frac{3(1+\sqrt{2})}{2}\;c=\frac{3(1-\sqrt{2})}{2},\;c=3,\;k=\frac{27}{4}$$

Yeah! It seems to me now that this is an easy problem! Thank you enormously, caffeinemachine! Last edited:

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Thank you enormously, caffeinemachine! Umm.. err.. uh.. You're.. err.. welcome. ##### Well-known member
I proceed slight differently

We have sum of the roots
a+b+c = 6

so (a + b+ c- 6) = 0
or (a-1)+(b-2)+(c-3) = 0

so (a-1)^3 + (b-2)^3 + (c-3)^3= 3(a-1)(b-2)(c-3) ( as x+ y + z = 0 => x^3+y^3 + z^3 = 3xyz

so 3(a-1)(b-2)(c-3) = 0 => a= 1 or b= 2 or c = 3

f(x) = x^3−6x^2+kx+k

taking 1 as a root we get

f(x)= 1-6+ k+ k = 0 or k = 5/2

so we get x^3-6x^2 + 5/2 x + 5/2 = 0

or 2x^3 – 12x^2 + 5 x + 5 = 0

factoring we get (x-1)(2x^2 – 10 x - 5) = 0

we can solve (2x^2 – 10 x - 5)= 0 to get (10+/-√(140))/4 so b = (5+ √35)/2 and c = (5 - √35)/2

another solution b = 5- √35)/2 and c = (5 - √35)/2

similarly other 4 sets of 3 roots can be found
b = 2 shall give 2 sets and c = 3 shall give 2 sets

( kindly note that as we have been given the condition

(b = x1, c= x2) is different from ( b = x2, c= x1))

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